Haskell

Managed to do part1 in one line using ByteString operations:

import Control.Monad
import qualified Data.ByteString.Char8 as BS

part1 :: IO Int
part1 =
  sum
    . ( BS.transpose . BS.split '\n'
          >=> fmap succ
          . BS.elemIndices 'O' . BS.reverse . BS.intercalate "#"
          . fmap (BS.reverse . BS.sort) . BS.split '#'
      )
    <$> BS.readFile "inp"

Part 2

{-# LANGUAGE NumericUnderscores #-}

import qualified Data.ByteString.Char8 as BS
import qualified Data.Map as M
import Relude

type Problem = [ByteString]

-- We apply rotation so that north is to the right, this makes
-- all computations easier since we can just sort the rows.
parse :: ByteString -> Problem
parse = rotate . BS.split '\n'

count :: Problem -> [[Int]]
count = fmap (fmap succ . BS.elemIndices 'O')

rotate, move, rotMov, doCycle :: Problem -> Problem
rotate = fmap BS.reverse . BS.transpose
move = fmap (BS.intercalate "#" . fmap BS.sort . BS.split '#')
rotMov = rotate . move
doCycle = rotMov . rotMov . rotMov . rotMov

doNcycles :: Int -> Problem -> Problem
doNcycles n = foldl' (.) id (replicate n doCycle)

findCycle :: Problem -> (Int, Int)
findCycle = go 0 M.empty
  where
    go :: Int -> M.Map Problem Int -> Problem -> (Int, Int)
    go n m p =
      let p' = doCycle p
       in case M.lookup p' m of
            Just n' -> (n', n + 1)
            Nothing -> go (n + 1) (M.insert p' n m) p'

part1, part2 :: ByteString -> Int
part1 = sum . join . count . move . parse
part2 input =
  let n = 1_000_000_000
      p = parse input
      (s, r) = findCycle p
      numRots = s + ((n - s) `mod` (r - s - 1))
   in sum . join . count $ doNcycles numRots p